Question

In: Statistics and Probability

A random sample of 45 restaurant servers found the average tips per shift were $150. Assume...

A random sample of 45 restaurant servers found the average tips per shift were $150. Assume the sample standard deviation is $25. Using a 99% confidence level, calculate the following: Confidence Interval Group of answer choices

A. (139.96; 160.04)

B. (141.76; 161.12)

C (140.09; 162.28)

D. (143.56; 167.90)

The average percentage of cities in the US that have snow on Christmas Day is 35%. A recent sample of 150 cities found that 21 had snow on Christmas Day last year. Using a 98% confidence level, calculate the following: Critical Value Group of answer choices

A. 2.33

B. 2.45

C. 1.96

D. 1.645

The average percentage of cities in the US that have snow on Christmas Day is 35%. A recent sample of 150 cities found that 21 had snow on Christmas Day last year. Using a 98% confidence level, calculate the following: standard error of the mean Group of answer choices

A. 03

B. .01

C. .02

D. .05

The average percentage of cities in the US that have snow on Christmas Day is 35%. A recent sample of 150 cities found that 21 had snow on Christmas Day last year. Using a 98% confidence level, calculate the following: Confidence interval Group of answer choices

A. (.07; .21)

B. (.09; .25)

C. (.08; .78)

D. (.09; .22)

Expert Solution

1)

option A is right.

2) for confidence interval for proportion we use Z distribution

for 98% confidence intwo decimal place this critical value is 2.33

option A is right.

3) Standard error = option A is right

4) confidence interval is

option A is right

orchestra answered 2 years ago

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